Computing work in an isothermal process

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In an isothermal process, the work done is calculated by integrating PdV from the initial to final volume, but pressure (P) is not constant along the curve. Instead, P is treated as a function of volume, temperature, and the number of particles, as described by the ideal gas law. During the process, while volume increases, pressure adjusts to maintain a constant temperature, distinguishing it from an isobaric process. If pressure were held constant, temperature would change with volume variations. Understanding these relationships is crucial for accurately computing work in thermodynamic cycles like the Carnot cycle.
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I was reviewing some notes on the Carnot cycle and noticed that to compute the work done along an isotherm you take the integral of PdV from the initial to final volume. What value do you use for P since P is not held constant on an isothermal curve? Why is P treated as a constant?
 
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If the gas being compressed in the Carnot engine is an ideal gas, then you may simply use the ideal gas law to express P in terms of temperature, particle number and volume:

P dV = nRT dV/V
 
But during that process,the volume is rising,so pressure remains constant.
 
R A V E N said:
But during that process,the volume is rising,so pressure remains constant.

It's an isothermal process being considered, not an isobaric process - the temperature (in an ideal gas) remains constant because the pressure changes to compensate for the change in volume. If the pressure were constant the temperature would have to change as the volume changed.
 

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